I want to prove this lemma which is left as exercise for the reader in the book Measure Theory and Integration of Micheal E.Taylor. The lemma is used to prove the Besicovitch Lemma. This is the statement:
$\mathbf{Claim}:$ There exists $K_{1}(n)$ such that if $r \in (0,+\infty),$ $B$ is a closed ball of radius $r,$ and $B_{1},\dots,B_{K_{1}(n)}$ are closed balls of radius $ \geq \frac{2}{3}r$ such that $B \cap B_{j} \neq \emptyset,$ then some $B_{j}$ contains the center of $B_{k}$ (for some $k \neq j$).
Until now, I have been able to prove only the statement with the following additional hypothesis: each of the radii is limited by a common constant $R_{0}$. I could stop here because a hypothesis of Besicovitch's lemma is the uniform boundness of the radii of the covering. Despite this, I would be interested in proving the claim above without this further hypothesis or I would like a counterexample.
I include below my proof of the $\mathbf{Claim}$ with the hypothesis of radii limited by a common constant $R_{0}.$
Suppose that $B_{1}, \dots, B_{K}$ are balls such that $B \cap B_{j} \neq \emptyset $ for every $j = 1, \dots, \, K$, and that no $B_{j}$ contains the center of another $B_{i}.$ The proof aims to find a bound on $K$ depending only on $n$ and $R_{0}.$ Firstly, indicate with $x_j$ and $r_{j}$ the center and the radius of $B_{j},$ and indicate $x_{B}$ the center of B. Since $B_{j}$ does not include the center of any other $B_{i}$ we obtain that $B(x_{j},\frac{1}{2}r_{j})$ and $B(x_{i},\frac{r_{i}}{2})$ are disjoint for every $i \neq j.$ In addition, there exists $C$ depending from $R_{0}$ such that $$ B(x_{j},\,\frac{1}{2}) \subseteq B(x_{B},\, C\cdot r) \quad \forall j. $$ But the balls $B(x_{j},\,\frac{r_{j}}{2})$ are all disjointed from each other, thus we have $$ \sum_{j=1}^{n} m\left(B(x_{j},\,\frac{1}{2}r_{j})\right) = m\left( \bigcup_{j=1}^{k} B(x_{j},\,\frac{1}{2}r_{j})\right) \leq m \left(B(x_{B}, C \cdot r)\right)= C^{2} \cdot \omega_{n} \cdot r^{n}, $$ where $m$ is the Lebesgue measure and $\omega_{n}$ the volume of the unitary ball. But all the radii $r_{j}$ are at least $\frac{2}{3}r$ and thus we obtain $$ K\cdot \frac{r^{n}}{3^{n}} \cdot \omega_{n} =\sum_{j=1}^{n} m\left(B(x_{j},\,\frac{1}{3}r)\right) \leq \sum_{j=1}^{n} m\left(B(x_{j},\,\frac{1}{2}r_{j}\right)) \leq C^{n} \cdot \omega_{n} \cdot r^{n}. $$ In conclusion, $K \leq 3^{n} \cdot C^{n}.$ Thus each set of closed balls intersecting $B$ having a radius of at least $2/3$ that of $B$, and having more than $3^{n} \cdot C^{n}$ must include a ball that contains the center of another ball in the set.